"Show that y=23ex+e−2x is a solution of the..." was answered by our community

Zaiden Soto

Open question

2022-08-20

Show that

y = \frac{2}{3} e^{x} + e^{- 2 x}

is a solution of the differential equation

y^{'} + 2 y = 2 e^{x}

Karsyn Lam

Beginner2022-08-21Added 7 answers

Step 1

y = \frac{2}{3} e^{x} + e^{- 2 x}

Differentiate with respect to x

y^{'} = \frac{2}{3} e^{x} - 2 e^{- 2 x}

Step 2
We have to verify the following differential equation:

y^{'} + 2 y = 2 e^{x}

Substitute the expressions for y' and y, To get

[\frac{2}{3} e^{x} - 2 e^{- 2 x}] + 2 [\frac{2}{3} e^{x} + e^{- 2 x}] = 2 e^{x}

[\frac{2}{3} e^{x} - 2 e^{- 2 x}] + [\frac{4}{3} e^{x} + 2 e^{- 2 x}] = 2 e^{x}

\frac{2}{3} e^{x} - 2 e^{- 2 x} + \frac{4}{3} e^{x} + 2 e^{- 2 x} = 2 e^{x}

(\frac{2}{3} + \frac{4}{3}) \cdot e^{x} = 2 e^{x}

Hence verified
Result:
Hint:

y^{'} = \frac{2}{3} e^{x} - 2 e^{- 2 x}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?